1. Quadratic Graphs - Parabolas
1. To draw a quadratic graph from y = ax2 + bx + c
2. Make up a table of values
3. Draw a smooth curve through the points plotted
4. The shape created is called a parabola

2. What do you think will be the values of y?
Quadratic Graphs
y = x2
Note: - X - = +
So -22 = -2 X -2 = 4
Use it to find the values of x2 x -2 -1 1 2 x2 y 4 1 1 4
What do you think will be the values of y?

3. Quadratic Graphs y = x2 x -2 -1 1 2 x2 4 y1 2 x2 4 y
Now join the points with a smooth curve.
y = x2
We need to plot the values (coordinates) of x and y on the grid to draw the graph.
What happens if y = x2 + 1 or y = x2 – 1.
Let’s Find out.

4. Quadratic Graphs y = x2 y = x2+1 y = x2 - 1
Note: - x - = +
So (-2)2 = -2 x -2 = 4
y = x2
y = x2+1
y = x2 - 1 x -2 -1 1 2 x2 4 y
Remember x2 is the same as before. All we need is to add 1 to x2 to find y. x -2 -1 1 2 x2 4
x2+1 y
Let’s do the same for when y = x2 - 1 x -2 -1 1 2 x2 4
x2-1 3 y
Let’s plot all the three graphs on the grid

5. Quadratic Graphs y = x2 y = x2 + 1 y = x2 - 1 x -2 -1 1 2 x2 4 y x -21 2 x2 4 y
y = x2 +1
y = x2 x -2 -1 1 2 x2 4
x2+1 5 y
y = x2 - 1 x -2 -1 1 2 x2 4
x2-1 3 y
What do you notice curves?
What do you notice about where the
lines cross the y-axis?

6. Quadratic Graphs y = x2 + 2x - 2
Remember that -2 in the equation is the intercept so it will run through x -3 -2 -1 1 2 x2 2x y

7. Now let’s multiply each value of x by 2 to find the values of 2x
Quadratic Graphs
Note: - X - = +
So -22 = -2 X -2 = 4
Use it to find the values of x2
y = x2 + 2x - 2 x -3 -2 -1 1 2 x2 2x y 9 4 1 1 4 -6 -4 -2 2 4
Now let’s multiply each value of x by 2 to find the values of 2x
+ X - = -
so 2 X -3 =-6

8. Quadratic Graphs y = x2 + 2x - 2
To find the values of y, we must sum all the values of x2, 2x and -2.
Be careful with the negative signs. x -3 -2 -1 1 2 x2 9 4 2x -6 -4 y 1 -2 -3 -2 1 6

9. Quadratic Graphs y = x2 + 2x - 2 x -3 -2 -1 1 2 x2 9 4 2x -6 -4 y 61 2 x2 9 4 2x -6 -4 y 6
y = x2 + 2x - 2
Plot the values of x and y on the grid and join the points with smooth curve.

10. Quadratic Graphs More negative values
y = x2 – 3x - 2 x -1 1 2 3 4 x2
-3x -2 y
Remember that -2 in the equation is the intercept so it will run through
Note: - X - = +
So -12 = -1 X -1 = 1
Use it to find the values of x2. These are just the ordinary square numbers.

11. Quadratic Graphs More negative values
y = x2 – 3x - 2 x -1 1 2 3 4 x2 9 16
-3x -2 y
Multiply each value of x by -3.
- X - = +
- X + = - 3 -3 -6 -9
-12 2 -2 -4 -4 -2 2
To find the values of y, we must sum all the values of x2, 2x and -2.
Be careful with negative signs. .

12. Quadratic Graphs More negative values
y = x2 – 3x - 2 x -1 1 2 3 4 x2 9 16
-3x -3 -6 -9
-12 -2 y -4
y = x2 – 3x - 2
Plot the values of x and y on the grid and join the points with smooth curve

13. y = x2 + 2x -15 x -6 -5 -4 -3 -2 -1 1 2 3 4 y 7 -7
-12
-15
-16

14. y = 12 – x - x2 x -5 -4 -3 -2 -1 1 2 3 4 y -8 6 10 12

15. Quadratic Equations
All quadratic expressions involve x2: y = ax2 + bx + c
The graphs of Quadratics are always parabolas
x2 negative
x2 positive
The points at which these graphs cut the x-axis are called the solution or roots of the equation

16. Solving Quadratic Equations from Graphs
y = x2 – 4x
= x(x – 4)
x(x – 4) = 0
x = 0 or x = 4
Roots of the equation

17. Solving Quadratic Equations from Graphs
y = x2 + 2x - 3
= (x + 3)(x – 1)
(x + 3)(x – 1) = 0
x = -3 or x = 1
Roots of the equation

18. Solving Quadratic Equations from Graphs
y = x - x2
= (8 + x)(3 – x)
(8 + x)(3 – x) = 0
x = -8 or x = 3
Roots of the equation

19. Solving Quadratic Equations from Graphs
y = 5 - 3x - 2x2
= (5 + 2x)(1 - x)
Roots of the equation
(5 + 2x)(1 - x) = 0
x = -2·5 or x = 1

20. Factorise and Solve ( find the roots of)
Mixture of examples

21. Roots When a number is multiplied by 0 the result is 0.
If 7b = 0 then b = 0
If ab = 0 then a = 0 or b = 0
If (x - a)(x - b) = 0 then either
(x - a) = 0 or (x - b) = 0
x = a or x = b
(x – 7)(x + 4) = 0
3x(x + 4) = 0
x – 7 = 0 or x + 4 = 0
3x = 0 or x + 4 = 0
x = 7 or x = -4
x = 0 or x = -4

22. Mixture of examples Set 1
1. 3x2 – 7x = 0
2. 4x2 – 9 = 0
x(3x – 7) = 0
(2x + 3)(2x – 3) = 0
x = 0 or 3x – 7 = 0
2x + 3 = 0 or 2x – 3 = 0
x = 0 or x = 7/3
x = - 3/2 or x = 3/2
3. x2 - 7x + 10 = 0
4. x2 - 2x - 48 = 0
(x - 8)(x + 6) = 0
(x - 2)(x – 5) = 0
x – 8 = 0 or x + 6 = 0
x – 2 = 0 or x – 5 = 0
x = 2 or x = 5
x = 8 or x = -6

23. In each example factorise then solve the equation.
1. 5x2 – 10x = 0
2. x2 – 16 = 0
5x(x – 2) = 0
x = 0, 2
(x + 4)(x – 4) = 0
x = -4, 4
3. x2 - 7x - 18 = 0
4. 2x2 – 3x = 0
x = 0, 3/2
x(2x – 3) = 0
(x - 9)(x + 2) = 0
x = -2, 9
6. 6x2 + 5x = 0
x2 – 49 = 0
x = 0, -5/6
(10x - 7)(10x + 7) = 0
x = -0·7, 0·7
x(6x + 5) = 0
7. x2 + 6x - 16 = 0
8. x2 + 5x + 6 = 0
(x - 2)(x + 8) = 0
x = -8, 2
x = -3, -2
(x + 2)(x + 3) = 0
9. 9x2 – 4 = 0
x2 – 2x = 0
(3x - 2)(3x + 2) = 0
x = -2/3, 2/3
x = 0, 1/4
2x(4x - 1) = 0
11. x2 - 6x + 8 = 0
12. x2 - x = 0
(x - 2)(x - 4) = 0
x = 2, 4
(x - 11)(x + 10) = 0
x = -10, 11

24. quadratic formula. If ax2 + bx + c = 0 then
When we cannot factorise or solve
quadratic equations graphically we need to use the
quadratic formula.
If ax2 + bx + c = 0
then

25. Example : Solve x2 + 3x – 3 = 0 ax2 + bx + c = 0 1 3 -3
= 0·79 or – 3·8

26. These are the roots of the equation.
Example 2
Use the quadratic formula to solve the equation :
x2 + 5x + 6= 0
ax 2 + bx + c= 0 1 5 6
x = - 2 or x = - 3
These are the roots of the equation.

27. These are the roots of the equation.
Example 3
Use the quadratic formula to solve the equation :
8x2 + 2x - 3= 0
ax 2 + bx + c= 0 8 2 -3
x = 1/2 or x = - 3/4
These are the roots of the equation.

28. These are the roots of the equation.
Example 4
Use the quadratic formula to solve the equation :
8x x + 15 = 0
ax 2 + bx + c= 0 8
-22 15
x = 3/2 or x = 5/4
These are the roots of the equation.

29. These are the roots of the equation.
Example 5
Use the quadratic formula to solve the equation :
2x 2 + 3x - 7 = 0
ax 2 + bx + c= 0 2 3 -7
x = 1·27 or x = -2·77
These are the roots of the equation.

30. x = -1·7, -0·3 x = -3·6, 0·6 x = 1·4, 0·4 x = 1·9, -0·9
Use quadratic formula to solve the following :
2x2 + 4x + 1 = 0
x2 + 3x – 2 = 0
x = -1·7, -0·3
x = -3·6, 0·6
5x2 - 9x + 3 = 0
3x2 - 3x – 5 = 0
x = 1·4, 0·4
x = 1·9, -0·9
4x2 - x - 1 = 0
x2 + 2x - 5 = 0
x = -0·39, 0·64
x = 1·45, -3·45

31. 67 Consider this rectangle
x + 7 x 67
The area of the rectangle is given by A = x(x + 7)
So we now obtain the equation x(x+7) = 67
This simplifies to give x2 + 7x - 67 = 0 , on moving all the terms to one side of the equation.

32. To solve this equation we use the quadratic equation
So we have a = 1 , b = 7 and c = - 67.
This gives

33. Which leads to So Giving the two solutions x = 5·40 and -12·40
This simplifies to
Which leads to So
Giving the two solutions x = 5·40 and -12·40

34. as you cannot have a negative length of side
Now x ≠
as you cannot have a negative length of side
so we are left with the solution that x = 5·40

35. Find x in each of the examples below.93
x + 2
x + 4 37

36. The roots of ax2 + bx + c = 0 are
Quadratic formula mixed examples
The roots of ax2 + bx + c = 0 are
1. Solve x2 - x – 5 = 0 , giving the roots correct to 1 decimal place.
2. Solve 2x2 – 5x + 1, giving the roots correct to 2 decimal places.
3. Solve 2x(x - 1) = 7 giving the roots correct to 1 decimal place.
The area of the rectangle shown is 43 sq.cm.
Find x correct to 1 decimal place.
x + 5
x + 1

37. 5. Use Pythagoras to form an equation in x and then solve it.
The triangle is right angled.
x + 5
x + 3 x
6. A quadratic equation shown has equation y = x2 – 8x + 11.
Find the coordinates of the points where
the curve cuts the x axis.
7. The height h of the object above the ground
after a time t seconds is given by
h = 15t – t2
When is the object 9 metres above the ground ?
Explain the meaning of the two answers.

38. Quadratics Cutting the x and y axis.
In each of the example:
a) Find the points where the quadratic
cuts the x and the y axis;
b) Use symmetry to determine the coordinates of the turning point .

39. Graph cuts the x-axis at (-8, 0) and (3, 0)
Ex 1. Sketch the graph of y = x2 + 5x - 24
Will the graph be “happy” or “sad”
x2 positive, so “happy”
Solve x2 + 5x – 24 = 0
The graph cuts the x-axis when y = 0
Now factorise
(x + 8)(x – 3) = 0
Solving gives
x + 8 = 0 or x – 3 = 0
x = or x = 3
Graph cuts y-axis at x = 0
Graph cuts the x-axis at (-8, 0) and (3, 0)
y = 02 +5(0) - 24
Cuts y axis at (0, -24)
By symmetry the x co-ordinate of turning point is midpoint of the roots -8 and 3.
x = (-8 + 3) ÷ 2 = -2·5
When x = 2·5, y = 2·52 + 5(-2·5) – 24 = 30·25
Turning point = (-2·5, -30·25)

40. Graph cuts the x-axis at (-5, 0) and (3, 0)
Ex 1. Sketch the graph of y = 15 – 2x - x2
Will the graph be “happy” or “sad”
x2 negative, so “sad”
Solve 15 – 2x – x2 = 0
The graph cuts the x-axis when y = 0
Same as x2 + 2x – 15 = 0
Now factorise
(x + 5)(x – 3) = 0
x + 5 = 0 or x – 3 = 0
Solving gives
x = or x = 3
Graph cuts y-axis at x = 0
Graph cuts the x-axis at (-5, 0) and (3, 0)
y = 15
Cuts y axis at (0, 15)
By symmetry the x co-ordinate of turning point is midpoint of the roots -5 and 3.
x = (-5 + 3) ÷ 2 = -1
When x = -1, y = (-1) – (-1)2 = 16
Turning point = (-1, 16)

41. Find the points where the quadratic cuts the x and the y axis.
Use symmetry to determine the coordinates of the turning point. D
y = x2 – 5x + 4 C B A y x 1.
(0,4)
(4,0)
(1,0)
(2·5,-2·25) D C B A y
y = x2 – 3x - 10 x 2.
x2 – 5x + 4 = 0
(x – 1)(x – 4) = 0
x = 1 or x = 4
(1 + 4) ÷ 2 = 2·5
y = 2·52 – 5(2·5) + 4 = -2·25
(2·5,-2·25)

42. x
y = x2 + 7x + 10 D C B A y 3. y
y = x2 – 8x + 7 4. D A B x C

43. y = x2 - 9 C B A x y 5. 6.
y = x2 – 6x B A y x

44. x
y = x2 – 4 x - 21 D C B A y 7. D B A y C x
y = x2 – 6x + 8 8.

45. D C B A
y = x2 + 7x - 8 y 9. x
y = x2 + 4 x + 3 D C B A y x
10.

46. Find the points where the quadratic cuts the x and the y axis.
Use symmetry to determine the coordinates of the turning point. 1. y x C B A
y = 16 – x2 y D C x B
y = 4 + 3x – x2 A 2.

47. Find the points where the quadratic cuts the x and the y axis.
Use symmetry to determine the coordinates of the turning point. 3. y D C x B
y = 5 – 4x – x2 A y x C B A
y = x – x2 4.

48. Find the points where the quadratic cuts the x and the y axis.
Use symmetry to determine the coordinates of the turning point. y D C x B
y = -14 – 9x – x2 A 5. y x C B A
y = 2 + x – x2 D 6.

49. Find the points where the quadratic cuts the x and the y axis.
Use symmetry to determine the coordinates of the turning point. y C B A
y = 49 – 4x2 7. y D B A
y = x – x2 C 8.

50. Find the points where the quadratic cuts the x and the y axis.
Use symmetry to determine the coordinates of the turning point. 9. y D C x B
y = 10 – 3x – x2 A y B A
y = -5x – x2 x
10. y B A
y = 7x – x2 x
11.